首页 技术 正文
技术 2022年11月23日
0 收藏 308 点赞 3,521 浏览 2396 个字

Out of Hay

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 13094 Accepted: 5078

Description

The cows have run out of hay, a horrible event that must be remedied immediately. Bessie intends to visit the other farms to survey their hay situation. There are N (2 <= N <= 2,000) farms (numbered 1..N); Bessie starts at Farm 1. She’ll traverse some or all of the M (1 <= M <= 10,000) two-way roads whose length does not exceed 1,000,000,000 that connect the farms. Some farms may be multiply connected with different length roads. All farms are connected one way or another to Farm 1.

Bessie is trying to decide how large a waterskin she will need. She knows that she needs one ounce of water for each unit of length of a road. Since she can get more water at each farm, she’s only concerned about the length of the longest road. Of course, she plans her route between farms such that she minimizes the amount of water she must carry.

Help Bessie know the largest amount of water she will ever have to carry: what is the length of longest road she’ll have to travel between any two farms, presuming she chooses routes that minimize that number? This means, of course, that she might backtrack over a road in order to minimize the length of the longest road she’ll have to traverse.

Input

* Line 1: Two space-separated integers, N and M.

  • Lines 2..1+M: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, describing a road from A_i to B_i of length L_i.

Output

* Line 1: A single integer that is the length of the longest road required to be traversed.

Sample Input

3 3

1 2 23

2 3 1000

1 3 43

Sample Output

43

Hint

OUTPUT DETAILS:

In order to reach farm 2, Bessie travels along a road of length 23. To reach farm 3, Bessie travels along a road of length 43. With capacity 43, she can travel along these roads provided that she refills her tank to maximum capacity before she starts down a road.

Source

USACO 2005 March Silver

求最小生成树的最大边,Kruskal

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <string>
#include <queue>
#include <vector>
#include <algorithm>
#define LL long long
using namespace std;const int INF = 0x3f3f3f3f;const int MAX = 11000;int n,m;int pre[3000];typedef struct node
{
int x;
int y;
int dis;
}K;
K Edge[MAX];
void init()
{
for(int i=1;i<=n;i++)
{
pre[i]=i;
}
}
bool cmp(node a,node b)
{
return a.dis<b.dis;
}int Find(int x)
{
return x==pre[x]?x:pre[x]=Find(pre[x]);
}int Kruskal()
{
int num=0,Max=0;
for(int i=0;i<m;i++)
{
int a=Find(Edge[i].x);
int b=Find(Edge[i].y);
if(a!=b)
{
num++;
pre[a]=b;
if(Max<Edge[i].dis)
{
Max=Edge[i].dis;
}
}
if(num==n-1)
{
break;
}
}
return Max;
}
int main()
{
while(~scanf("%d %d",&n,&m))
{
init();
for(int i=0;i<m;i++)
{
scanf("%d %d %d",&Edge[i].x,&Edge[i].y,&Edge[i].dis);
}
sort(Edge,Edge+m,cmp);
printf("%d\n",Kruskal());
}
return 0;
}

微信扫一扫

支付宝扫一扫

本文网址:https://www.zhankr.net/140654.html

相关推荐
python开发_常用的python模块及安装方法
adodb:我们领导推荐的数据库连接组件bsddb3:BerkeleyDB的连接组件Cheetah-1.0:我比较喜欢这个版本的cheeta…
日期:2022-11-24 点赞:877 阅读:5,772
Educational Codeforces Round 11 C. Hard Process 二分
C. Hard Process题目连接:http://www.codeforces.com/contest/660/problem/CDes…
日期:2022-11-24 点赞:806 阅读:3,859
下载Ubuntn 17.04 内核源代码
zengkefu@server1:/usr/src$ uname -aLinux server1 4.10.0-19-generic #21…
日期:2022-11-24 点赞:565 阅读:4,665
可用Active Desktop Calendar V7.86 注册码序列号
可用Active Desktop Calendar V7.86 注册码序列号Name: www.greendown.cn Code: &nb…
日期:2022-11-24 点赞:731 阅读:4,608
Android调用系统相机、自定义相机、处理大图片
Android调用系统相机和自定义相机实例本博文主要是介绍了android上使用相机进行拍照并显示的两种方式,并且由于涉及到要把拍到的照片显…
日期:2022-11-24 点赞:512 阅读:5,699
Struts的使用
一、Struts2的获取  Struts的官方网站为:http://struts.apache.org/  下载完Struts2的jar包,…
日期:2022-11-24 点赞:671 阅读:3,401
发表评论
暂无评论

还没有评论呢,快来抢沙发~

助力内容变现

将您的收入提升到一个新的水平

点击联系客服

在线时间:8:00-16:00

客服电话

400-888-8888

客服邮箱

ceotheme@ceo.com

扫描二维码

关注微信公众号

扫描二维码

手机访问本站